Some Topological Properties of Charming Spaces

LI XIAO-TINGLIN FU-CAIAND LIN SHOU

(1.School of Mathematics and Statistics,Minnan Normal University, Zhangzhou,Fujian,363000)

(2.Institute of Mathematics,Ningde Teachers’College,Ningde,Fujian,352100)

Some Topological Properties of Charming Spaces

LI XIAO-TING1,LIN FU-CAI1,*AND LIN SHOU2

(1.School of Mathematics and Statistics,Minnan Normal University, Zhangzhou,Fujian,363000)

(2.Institute of Mathematics,Ningde Teachers’College,Ningde,Fujian,352100)

Communicated by Lei Feng-chun

In this paper,we mainly discuss the class of charming spaces.First,we show that there exists a charming space such that the Tychonof f product is not a charming space.Then we discuss some properties of charming spaces and give some characterizations of some class of charming spaces.Finally,we show that the Suslin number of an arbitrary charming rectif i able space is countable.

charming space,(i,j)-structured space,Lindel¨ofΣ-space,Suslin number,rectif i able space

1 Introduction

In 1969,Nagami[1]introduced the notion ofΣ-spaces,and then the class ofΣ-spaces with the Lindel¨of property(i.e.,the class of Lindel¨ofΣ-space)quickly attracted the attention of some topologists.From then on,the study of Lindel¨ofΣ-spaces has become an important part in the functional analysis,topological algebra and descriptive set theory.Tkachuk[2]described detailedly Lindel¨ofΣ-spaces and made an overview of the recent progress achieved in the study of Lindel¨ofΣ-spaces.Arhangel’skii[3]has proved if the weight ofXdoes not exceed 2ω,then any remainders ofXin a Hausdorf f compactif i cation is a Lindel¨ofΣ-space.

It is natural to ask if we can f i nd a class of spacesPsuch that each remainder of a Hausdorf f compactif i cation of arbitrary metrizable space belongs toP.Therefore,Arhangel’skiidef i ned charming spaces in[3]and showed the any remainder of paracompactp-space in a Hausdorf f compactif i cation is a charming space.Indeed,Arhangel’skii def i ned many new classes of spaces(that is,(i,j)-structured spaces)which have similar structure with the class of charming spaces,and he said that each of the classes of spaces so def i ned is worth studying. Therefore,we mainly discuss some topological properties of(i,j)-structured spaces.

2 Preliminaries

All spaces are Tychonof f unless stated otherwise.Readers may refer to[4]–[5]for notations and terminology not explicitly given here.

Def i nition 2.1Let N be a family of subsets of a space X.Then the family N is a network of X if every open subset U is the union of some subfamily of N.

Def i nition 2.2We say that a space is cosmic if it has a countable network.

Def i nition 2.3Let X be a space.We say that X is a Lindel¨of p-space if it is the preimage of a separable metrizable space under a perfect mapping.

Def i nition 2.4X is a Lindel¨of Σ-space if there exists a space Y which maps continuously onto X and perfectly onto a second countable space.

That is,a Lindel¨ofΣ-space is the continuous onto image of some Lindel¨ofp-space. Therefore,a Lindel¨ofp-space is a Lindel¨ofΣ-space.It is well-known that the class of Lindel¨ofΣ-spaces contains the classes ofΣ-compact spaces and spaces with a countable network.

Def i nition 2.5[3]Let X be a space.If there exists a Lindel¨of Σ-subspace Y such that for each open neighborhood U of Y in X we have XU is also a Lindel¨of Σ-subspace,then we say that X is a charming space.

Def i nition 2.6[3]Let P and Q be two classes of topological spaces respectively.A space X will be called(P,Q)-structured if there is a subspace Y of X such that Y∈P,and for each open neighborhood U of Y in X,the subspace XU of X belongs to Q.In this situation,we call Y a(P,Q)-shell of the space X.

Def i nition 2.7[3]Let P0be the class of Σ-compact spaces,P1be the class of separable metrizable spaces,P2be the class of spaces with a countable network,P3be the class of Lindel¨of p-spaces,P4be the class of Lindel¨of Σ-spaces and P5be the class of compact spaces.Choose some i,j∈{0,1,2,3,4,5}.A space X is called(i,j)-structured if it is(Pi,Pj)-structured.In particular,a(4,4)-structured space is called a charming space.

3 The Properties of(i,j)-structured Spaces

In this section,we discuss some properties of(i,j)-structured spaces.

Proposition 3.1Each closed subspace of an(i,j)-structured space X is(i,j)-structured, where i,j∈{0,1,2,3,4,5}.

Proof.LetXbe an(i,j)-structured space,Abe a closed subspace ofX.SinceXis an (i,j)-structured space,there is a subspaceYsuch thatYis a(Pi,Pj)-shell of the spaceX.LetB=A∩Y.We claim thatBis a(Pi,Pj)-shell of the spaceA.Obviously,we haveB∈Pi.Now it suffi ces to show that for each open neighborhoodVofBinA,the subspaceAVofAbelongs toPj.Indeed,letVbe an open neighborhood ofBinA.Then there is an open neighborhoodV1inXsuch that

We have

and

It is easy to see that

The following three propositions are easy exercises.

Proposition 3.2Any image of an(i,j)-structured space under a continuous mapping is an(i,j)-structured space,where i,j∈{0,2,4,5}.

Proposition 3.3Any preimage of an(i,j)-structured space under a perfect mapping is an(i,j)-structured space,where i,j∈{0,3,4,5}.

Question 3.1For arbitraryi,j∈{1,3},is any image of an(i,j)-structured space under a continuous mapping an(i,j)-structured space?

Question 3.2For arbitraryi,j∈{1,2},is any preimage of an(i,j)-structured space under a perfect mapping an(i,j)-structured space?

Therefore,Xis an(i,j)-structured space.

Since the intersection of countably many Lindel¨ofΣ-subspaces is also Lindel¨ofΣ,it is natural to pose the following question.

It is well-known that the product of a countably many Lindel¨ofΣ-spaces is also a Lindel¨ofΣ-space.However,the product of two charming spaces may not be a charming space,see Example 3.1.

We know that each discrete spaceXhas a Hausdorf f one point Lindel¨of i cation which def i ned as follows:take an arbitrary pointp/∈Xand consider the setY=X∪{p},and then let all points ofXbe open and each neighborhood ofpbe the formU∪{p},whereUis open inXandXUis a countable set.

Example 3.1There exists a charming spaceXsatisfying the following conditions:

(i)Xis not aΣ-space;

(ii)The product ofX2is not a charming space.

Proof.LetX={∞}∪Dbe the one-point Lindef i cation of an uncountable discrete space,whereDis an uncountable discrete space.Tkachuk[2]has proved thatXis not a Lindel¨ofΣ-space.Obviously,the subspace{∞}is aΣ-space,and for each open neighborhoodVof{∞}inXwe haveDVisthenDVis separable and metrizable.ThusDVis a Lindel¨ofΣ-space.Therefore,Xis a charming space.

Now,we shall show thatX2is not a charming space.Assume thatX2is a charming space. Then there exists a Lindel¨ofΣ-subspaceL⊂X2such that,for each open neighborhoodUofL,X2Uis a Lindel¨ofΣ-subspace.It is easy to see that(∞,∞)∈L.Since each compact subset ofX2is f i nite,Lmust be a countable set by Theorem 5 in[2].Then there exists a pointx0∈Dsuch that({x0}×X)∩L=∅and(X×{x0})∩L=∅.It is easy to see that ((X{x0})×(X{x0}))is an open neighborhood ofLinX2.PutV=X{x0}.Then

is a Lindel¨ofΣ-space.Since{x0}×Vis closed inX2(V×V),we see that{x0}×Vis a Lindel¨ofΣ-space.However,Vis homeomorphic toX,which is a contradiction.Therefore,X2is not a charming space.

Remark 3.1(1)Tkachuk[2]proved that ifXis a Lindel¨ofΣ-space and every compact subset ofXis f i nite thenXis countable.From Example 3.1,we know that we can not replace“Lindel¨ofΣ-spaces”by“charming spaces”.

(2)The product of a family of f i nitely many charming spaces need not be a charming space.

(3)The spaceXin Example 3.1 is(i,j)-structured fori,j∈{0,1,2,3,4}.Therefore,the product of a family of f i nitely many(i,j)-structured spaces need not be an(i,j)-structured space.

It is easy to check thatX2in Example 3.1 is a Lindel¨of space.Therefore,we have the following question:

Question 3.4Is the product of two charming spaces Lindel¨of?In particular,is the product of a charming space with a Lindel¨ofΣ-space a Lindel¨of space?

Finally,we discuss a charming space with aGδ-diagonal.

Theorem 3.1[2]A Lindel¨of Σ-space with a Gδ-diagonal has a countable network.

Proposition 3.6Let S be the Sorgenfrey line.Then any uncountable subspace of S is not a Lindel¨of Σ-subspace.

Proof.LetYbe an uncountable subspace ofS.Assume thatYis a Lindel¨ofΣ-subspace. ThenL=Y∪(−Y)is also a Lindel¨ofΣ-subspace ofS,henceL2has a countable network by Theorem 3.1 sinceShas aGδ-diagonal.However,L2contains a closed,uncountable and discrete subspace{(x,−x):x∈L},which is a contradiction sinceL2has a countable network.

Example 3.2 There exists a hereditarily Lindel¨of spaceXwhich is not a charming space.

Proof.LetXbe the Sorgenfrey line.ThenXis a hereditarily Lindel¨of space.Assume thatXis a charming space.Then there exists a subspaceYsuch thatYis a(P4,P4)-shell inX.By Proposition 3.6,Yis countable.LetY={bn:n∈N}.For eachn∈N,take an open neighborhood[bn,bn+2−n)inX.Then

is an open neighborhood ofYinX.Since

we have thatXUis an uncountable set.ThenXUis not a Lindel¨ofΣ-space by Proposition 3.6,which is a contradiction.

It is natural to ask the following question.

Question 3.5Does each charming spaceXwith aGδ-diagonal have a countable network?

By Theorem 3.1,the following proposition is obvious.

Proposition 3.7Each charming space X with a Gδ-diagonal is a(2,2)-structured space.

The following two results are well-known in the class of Lindel¨ofΣ-spaces.

Theorem 3.2[6]Each hereditarily Lindel¨of Σ-space has a countable network.

Theorem 3.3[5]Each Lindel¨of Σ-space with a point-countable base is second-countable.

However,the following questions are remain open.

Question 3.6Is each hereditarily charming space a Lindel¨ofΣ-space?

Question 3.7Does each hereditarily charming space have a countable network?

Question 3.8Is each charming space with a point-countable base metrizable?

4 CL-charming and CO-charming Spaces

In this section,we introduce two classes of charming spaces,and then discuss some properties of them.

Def i nition 4.1A space X is called CL-charming space(resp.,CO-charming space)if there is a closed subspace(resp.,compact subspace)Y of X such that Y is a(P4,P4)-shell in X.

Obviously,each CO-charming space is CL-charming,and each CL-charming is a charming space.The spaceXin Example 3.1 is CO-charming.However,there exists a CL-charming space is not CO-charming space.

Example 4.1 There exists a CL-charming space which is not CO-charming.

The following three propositions are easy to check.

Proposition 4.1Any image of a CO-charming space under a continuous mapping is a CO-charming space.

Proposition 4.2Any closed image of a CL-charming space under a continuous mapping is a CL-charming space.

Proposition 4.3Any preimage of a CL-charming space(resp.,CO-charming space)under a perfect mapping is a CL-charming space(resp.,CO-charming space).

However,we do not know the answer of the following question.

Question 4.1Is any image of a CL-charming space under a continuous mapping a CL-charming space?

Theorem 4.1A hereditarily CL-charming space X is a hereditarily Lindel¨of Σ-space. Therefore,X has a countable network.

Theorem 4.2A CO-charming space X with a point-countable base is metrizable.

Proof.LetBbe a point-countable base ofX.SinceXis CO-charming,there exists a compact subspaceKofXsuch thatKis a(P4,P4)-shell inX.Moreover,since a compact space with a point-countable base is metrizable,Kis a separable and metrizable space.PutB′={B∈B:B∩.ThenB′is countable sinceKis separable.Let

Then it is easy to check thatU={Un:n∈N}is a countable base ofKinX.For eachn∈N,letXn=XUn.Since eachXnis a Lindel¨ofΣ-subspace with a point-countable base,eachXnis separable and metrizable by Theorem 3.3.Therefore,

which implies thatXis separable.Since a separable space with a point-countable base has a countable base,Xis metrizable.

The following result gives a partial answer to Question 3.5.

Theorem 4.3A CO-charming space with a Gδ-diagonal has a countable network.

is a Lindel¨ofΣ-subspace.Therefore,is a Lindel¨ofΣ-space.By Theorem 4 in[2],Xhas a countable network.

5 The Characterizations of Weak(i,j)-structured Spaces

Def i nition 5.1For any i,j∈{0,1,2,3,4,5},a space X is called a weak(i,j)-structured space if there exists a space Y which maps continuously onto X and perfectly onto a(i,j)-structured space.

Obviously,each weak(1,1)-structured space is a charming space,and hence it is a Lindel¨of space.However,the following question is still open.

Question 5.1Is each charming space weak(1,1)-structured?

Proposition 5.1Each closed subspace F of a weak(i,j)-structured space X is also a weakly(i,j)-structured space.

Proof.Take a spaceYfor which there exists a continuous onto mapφ:Y→Xand a perfect maph:Y→Mfor some(i,j)-structured spaceM.LetFbe a closed subspace ofYand letZ=φ−1(F).ThenFis a continuous image ofZand it is easy to see thath|Z:Z→Mis a perfect map.HenceFis a weak(i,j)-structured space.

The following theorem gives a characterization of weak(i,j)-structured spaces.First, we recall some concepts.

Any mapφfrom a spaceXto the family exp{Y}of subsets ofYis called multivalued; for convenient,we always writeφ:X→Yinstead ofφ:X→exp{Y}.A multivalued mapφ:X→Yis called compact-valued(resp.,f i nite-valued)if the setφ(x)is compact(resp., fi nite)for eachx∈X.

Letφ:X→Ybe a multivalued map.For anyA⊂X,we denote byφ(A)the set∪{φ(x):x∈A},that is,φ(A)=∪{φ(x):x∈A};we say that the mapφis onto ifφ(X)=Y.The mapφis called upper semicontinuous ifφ−1(U)={x∈X:φ(x)⊂U}is open inXfor any open subsetUinY.

Theorem 5.1For arbitrary i,j∈{0,4,5},the following conditions are equivalent for any space X:

(1)X is a weak(i,j)-structured space;

(2)there exist spaces K and M such that K is compact,M is an(i,j)-structured space and X is a continuous images of a closed subspace of K×M;

(3)X belongs to any class P which satisf i es the following conditions:

(a)P contains compact spaces and(i,j)-structured spaces;

(b)P is invariant under closed subspaces and continuous images;

(c)P5×P is contained in P;

(4)there is an upper semicontinuous compact-valued onto a map φ:M→X for some(i,j)-structured space M.

Proof.(1)⇒(2).There exists a spaceYwhich maps continuously ontoXand perfectly onto an(i,j)-structured spaceM.Then we f i x the respective perfect maph:Y→M.Leti:Y→βYbe the identity embedding.Then the diagonal mapg=h∆i:Y→M×βYis perfect by Theorem 3.7.9 in[4],and so the setg(Y)is closed inM×βY.Sincegis injective,g(Y)is homeomorphic toY,and henceXis a continuous image ofg(Y).

(2)⇒(3).Suppose thatXis a continuous image of a closed subsetFof the productM×Kfor some(i,j)-structured spaceMand compact spaceK.Take any classPas in (2).ThenK∈PandM∈P.ThereforeM×K∈Pand henceF∈Pas well sincePis invariant under closed subspaces.SincePis invariant under continuous images,we haveX∈P.

(3)⇒(1).First,we show ifXsatisf i es(1)then any closed subspaceFofXalso satisf i es (1).Indeed,take a spaceYfor which there exists a continuous onto mapφ:Y→Xand a perfect maph:Y→Mfor some(i,j)-structured spaceM.LetFis an any closed subspace ofXand letZ=φ−1(F).ThenFis a continuous image ofZand it is easy to see thath|Z:Z→Mis a perfect map.HenceFsatisf i es(1).It is evident that the class of spaces satisf i es(1)is invariant under continuous images.Moreover,the classes of(i,j)-structured spaces and compact spaces satisfy(1).

(1)⇒(4).Suppose thatXsatisf i es(1),i.e.,there exists a spaceYwhich maps continuously ontoXand perfectly onto an(i,j)-structured spaceN.Fix the respective mapf:Y→Xand a perfect mapg:Y→N.LetM=g(Y).For everyx∈X,since the setφ(x)=f(g−1(x))⊂Xis compact,φ:M→Xis a compact-valued map.The mapfis surjective,and henceφ(M)=X.It is easy to see thatφis upper semicontinuous.

(4)⇒(2).Fix an(i,j)-structured spaceMand a compact-valued upper semicontinuous onto mapφ:M→X.PutF=∪{φ(x)×{x}:x∈M}.ThenFis contained inβX×M. Letπ:βX×M→βXbe the projection.Thenπ(F)=XsoXis a continuous image ofF.Fix any point(x,t)∈(βX×M)F.Thenx/∈φ(t),and hence we can f i nd disjoint setsU,V∈τ(βX)such thatx∈Uandφ(t)⊂V.Sinceφis upper semicontinuous,there exists a setW∈τ(t,M)such thatφ(W)⊂V.It easily check that(x,t)∈U×W⊂(βX×M)F,therefore the setFis closed inβX×M.

6 Rectif i able Spaces with a Weak(i,j)-structure

Recall that a topological groupGis a groupGwith a(Hausdorf f)topology such that the product maps ofG×GintoGis jointly continuous and the inverse map ofGonto itself associatingx−1with arbitraryx∈Gis continuous.A topological spaceGis said to be a rectif i able space provided that there are a surjective homeomorphismφ:G×G→G×Gand an elemente∈Gsuch thatπ1◦φ=π1and for everyx∈Gwe haveφ(x,x)=(x,e),whereπ1:G×G→Gis the projection to the f i rst coordinate.IfGis a rectif i able space,thenφis called a rectif i cation onG.It is well known that rectif i able spaces are a good generalizations of topological groups.In fact,for a topological group with the neutral elemente,then it is easy to see that the mapφ(x,y)=(x,x−1y)is a rectif i cation onG.However,the 7-dimensional sphereS7is rectif i able but not a topological group(see[8]).

Theorem 6.1[9]–[11]A topological space G is rectif i able if and only if there exist a e∈G and two continuous maps p:G2→G,q:G2→G such that for any x,y∈G the next identities hold:

Given a rectif i cationφof the spaceG,we may obtain the mappingspandqin Theorem 6.1 as follows.Letp=π2◦φ−1andq=π2◦φ.Then the mappingspandqsatisfy the identities in Theorem 6.1,and both are open mappings.

LetGbe a rectif i able space,andpbe the multiplication onG.Further,we sometimes writex·yinstead ofp(x,y)andA·Binstead ofp(A,B)for anyA,B⊂G.Therefore,q(x,y)is an element such thatx·q(x,y)=y,sincex·e=x·q(x,x)=xandx·q(x,e)=e, it follows thateis a right neutral element forGandq(x,e)is a right inverse forx.Hence a rectif i able spaceGis a topological algebraic system with operationspandq,a 0-ary operatione,and identities as above.It is easy to see that this algebraic system need not to satisfy the associative law about the multiplication operationp.Clearly,every topological loop is rectif i able.

Obviously,if A is a Lindel¨of Σ-subspace,then H is a Lindel¨of Σ-space.

Proof.Sinceq(A,A)⊂B1,we havee∈B1.Therefore,it is easy to see that

Put

Next we shall prove thatHis a rectif i able subspace ofG.Indeed,take arbitrary pointsx,y∈B.Then there exists ann∈N such thatx,y∈An∪Bn,and hence

Therefore,His a rectif i able subspace ofG.

Since the product of a countable family of Lindel¨ofΣ-space is Lindel¨ofΣ,it is easy to see thatHis Lindel¨ofΣ.

Theorem 6.2Every charming rectif i able space G has a dense rectif i able subspace that is a Lindel¨of Σ-space.

Proof.SinceGis charming,there exists a subspaceBsuch thatGis a(P4,P4)-shell ofG.

Case 1.Bis dense inG.

Take the smallest rectif i able subspaceHofGsuch thatL⊂G.ThenHis also a Lindel¨ofΣ-space by Lemma 6.1.Clearly,His dense inG.

Case 2.Bis not dense inG.

Then there exists a non-empty open subsetUofGsuch thatis an open neighborhood ofB,and henceis a Lindel¨ofΣ-space.Thenis an open cover ofG,and each element ofAis homeomorphic to.SinceGis Lindel¨of, there exists a countable subcover ofA,and henceGis a Lindel¨ofΣ-space.

A topological spaceXhas the Suslin property if every pairwise disjoint family of nonempty open subsets ofXis countable.

Lemma 6.2[12]The Suslin number of an arbitrary Lindel¨ofΣ-rectif i able space G is countable.

Theorem 6.3The Suslin number of an arbitrary charming rectif i able space G is countable.

Proof.By Theorem 6.2,Ghas a dense rectif i able subspaceHwhich is a Lindel¨ofΣ-subspace.Then the Suslin number ofHis countable by Lemma 6.2.SinceHis dense inG, it follows that the Suslin number ofGis countable.

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tion:54E20,54E35,54H11,22A05

A

1674-5647(2017)02-0110-11

10.13447/j.1674-5647.2017.02.02

Received date:Dec.6,2015.

Foundation item:The NSF(11571158,11471153 and 11201414)of China,the NSF(2017J01405,2016J05014, 2016J01671 and 2016J01672)of Fujian Province of China.

*Corresponding author.

E-mail address:985513859@qq.com(Li X T),linfucai@mnnu.edu.cn(Lin F C).